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If $$x^4-x^3-13x^2+26x-8 = (x-a)(x-b)(x-c)(x-d)$$ Such that$$cd=-8\\a>b\\c<d$$ What are $a,b,c$ and $d$?


Since the problem gave us the polynomial, I thought we can just expand the $(x-a)(x-b)(x-c)(x-d)$ out and match the coefficients, it turned out to be: $$abcd=-8\\a+b+c+d=1\\ab+ac+ad+bc+bd+cd=-13\\abc+acd+abd+bcd=-26$$ Can I solve $a,b,c,d$ from there? How can I do so?

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  • $\begingroup$ Note that $ab=1$. Along with the given fact that $a \gt b$ you know two roots are not integers. By the rational root theorem they are irrational. $\endgroup$ – Ross Millikan Jan 28 at 4:32
  • $\begingroup$ Watch your signs. In your last equation you have $26$ where you should have $-26.$ $\endgroup$ – David K Jan 28 at 4:43
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You know that $abcd = -8$ and $cd = -8.$ Therefore $ab= 1.$

From the coefficient of $x$ in the polynomial, you know that $abc + acd + abd + bcd = -26$ (note: not $+26$).

But since $ab=1$ and $cd=-8$, you can see that $abc + acd + abd + bcd = c - 8a + d - 8b.$ Therefore

$$ -8a - 8b + c + d = -26. \tag1 $$

But you also know that

$$ a + b + c + d = 1. \tag2 $$

Subtract Equation $(1)$ from Equation $(2)$:

$$ 9a + 9b = 27. $$

That is, $a + b = 3.$ But $b = \frac1a,$ so we have

\begin{align} a + \frac1a &= 3, \\ a^2 + 1 &= 3a, \\ a^2 - 3a + 1 &= 0. \\ \end{align}

Apply the quadratic formula to solve $y^2 - 3y + 1 = 0.$ Note by symmetry that $a$ and $b$ both are solutions of this equation. But you are given that $a > b$, so you can see how to match $a$ and $b$ with the two solutions of the quadratic formula.

For $c$ and $d,$ multiply Equation $(2)$ by $8$ and add the result to Equation $(1)$. You get $9c + 9d = -18.$ But also $d = -\frac8c.$ Again you can get a quadratic equation out of this and solve it, then use the information that $c < d$ to know which root is which.

No guesswork is required, although if you do guess cleverly you can shorten the path a little.

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Let $$f(x) = x^4-x^3-13x^2+26x-8$$

Observe that $f(2) = 0$ and $f(-4) = 0$

Hence $(x-2)$ and $(x+4)$ are two factors of $f(x)$

Also $f(x)$ is divisible by $(x-2)(x+4) = x^2 + 2x - 8$

Now divide $f(x)$ by $x^2 + 2x - 8$ to obtain the other quadratic factor $x^2 - 3x +1$

If you solve $x^2 - 3x +1 = 0$, you'll find $x = \frac{3}{2} \pm \frac{\sqrt{5}}{2}$

Finally $$f(x) = \left(x - \left(\frac{3}{2} + \frac{\sqrt{5}}{2}\right) \right) \left(x - \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right) \right) (x + 4)(x-2)$$

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    $\begingroup$ How did you observe that? $\endgroup$ – Cyh1368 Jan 28 at 4:22
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    $\begingroup$ Consider the factors of $-8$ $\endgroup$ – PTDS Jan 28 at 4:25
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    $\begingroup$ @Cyh1368: the rational root theorem tells you that any rational roots are among $\pm1, \pm2, \pm4, \pm8$. In a school problem you are pretty well guaranteed that you can get to a quadratic this way. $\endgroup$ – Ross Millikan Jan 28 at 4:31

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